A body moving with uniform acceleration covers a distance of 14 m in first 2 second and 88 m in next 4 second. How much distance is travelled by it in `5^(th)` second?

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We know that a body is moving with uniform acceleration. We are given the distances covered in two intervals: 14 m in the first 2 seconds and 88 m in the next 4 seconds. We need to find the distance traveled in the 5th second.

### Step 2: Use the kinematic equation for the first interval
For the first 2 seconds, we can use the kinematic equation:
\[ S = Ut + \frac{1}{2}At^2 \]
Where:
- \( S \) = distance covered (14 m)
- \( U \) = initial velocity
- \( A \) = acceleration
- \( t \) = time (2 seconds)

Substituting the values, we get:
\[ 14 = U(2) + \frac{1}{2}A(2^2) \]
\[ 14 = 2U + 2A \]
This simplifies to:
\[ 2U + 2A = 14 \]
Dividing through by 2:
\[ U + A = 7 \quad \text{(Equation 1)} \]

### Step 3: Use the kinematic equation for the second interval
For the next 4 seconds (from 2 seconds to 6 seconds), the total distance covered is 88 m. The total distance from the start to 6 seconds is:
\[ S = U(6) + \frac{1}{2}A(6^2) \]
So, we have:
\[ 14 + 88 = U(6) + \frac{1}{2}A(36) \]
This simplifies to:
\[ 102 = 6U + 18A \]
Dividing through by 6:
\[ 17 = U + 3A \quad \text{(Equation 2)} \]

### Step 4: Solve the equations simultaneously
Now we have two equations:
1. \( U + A = 7 \)
2. \( U + 3A = 17 \)

Subtract Equation 1 from Equation 2:
\[ (U + 3A) - (U + A) = 17 - 7 \]
This simplifies to:
\[ 2A = 10 \]
Thus, we find:
\[ A = 5 \, \text{m/s}^2 \]

Now substitute \( A \) back into Equation 1:
\[ U + 5 = 7 \]
So,
\[ U = 2 \, \text{m/s} \]

### Step 5: Calculate the distance traveled in the 5th second
The distance covered in the nth second is given by the formula:
\[ S_n = U + \frac{1}{2}A(2n - 1) \]
For the 5th second (\( n = 5 \)):
\[ S_5 = 2 + \frac{1}{2}(5)(2 \times 5 - 1) \]
Calculating:
\[ S_5 = 2 + \frac{1}{2}(5)(9) \]
\[ S_5 = 2 + \frac{45}{2} \]
\[ S_5 = 2 + 22.5 = 24.5 \, \text{m} \]

### Final Answer
The distance traveled by the body in the 5th second is **24.5 meters**.

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